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A 0.5505 g sample of impure Ca(OH)2 is dissolved in 50.00 mL of water. 20.00 mL of the resulting solution is then titrated with 0.1939 M HCl. What is the percent purity of the Ca(OH)2 if the titration requires 22.18 mL of acid? g

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mickymike92

Based on the data provided, the percentage purity of Ca(OH)2 is 28%.

What is the mass of Ca(OH)2 that will be diluted by 22.18 mL of 0.1939 M HCl?

The equation of the reaction is given below:

  • Ca(OH)2 + 2 HCl ----> CaCl2 + 2H20
  • 2 moles of HCl neutralizes 1 mole of Ca(OH)2

Mole of HCl in 22.18 mL of 0.1939 M solution is calculated using the formula:

  • moles = molarity × volume in L

volume of HCl = 0.02218 L

moles of HCl = 0.02218 × 0.1939

moles of HCl = 0.0043 moles

Moles of Ca(OH)2 neutralized = 0.0043/2 = 0.00215 moles

mass of 0.00215 moles of Ca(OH)2 is calculated using the formula:

  • mass = moles × molar mass

molar mass of Ca(OH)2 = 74 g/mol

mass of Ca(OH)2 = 0.00215 × 74

mass of Ca(OH)2 = 0.1591

What is the percentage purity of the Ca(OH)2?

Percentage purity is calculated using the formula:

  • Percentage purity = mass of pure Ca(OH)2/mass of impure × 100%

Percentage purityof Ca(OH)2 = 0.1591/0.5505 × 100%

Percentage purity of Ca(OH)2 = 28%

Therefore, the percentage purity of Ca(OH)2 is 28%.

Learn more about percentage purity at: https://brainly.com/question/24512674

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